A 'complicated' integral: $ \int \limits_{-\infty}^{\infty}\frac{\sin(x)}{x}$

Question

I am calculating an integral $\displaystyle \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x}$ and I dont seem to be getting an answer.

When I integrate by parts twice, I get:
$$\displaystyle \int \limits _{-\infty}^{\infty}\frac{\sin(x)}{x}dx = \left[\frac{\sin(x)\ln(x) - \frac{\cos(x)}{x}}{2}\right ]_{-\infty}^{+\infty}$$

What will be the answer to that ?

You could use double integrals and switch limits. It is an improper integral. The answer is $\pi$. — Torsten Hĕrculĕ Cärlemän, Jul 17 '13 at 20:09
@AnuragPallaprolu I looked it up on Wolfram Alpha and it said $\pi$. WHat is an improper integral ? :) — An SO User, Jul 17 '13 at 20:11
@LittleChild I think we could use a bit of wiki here. :) Its an integral whose limits reach either infinities. — Torsten Hĕrculĕ Cärlemän, Jul 17 '13 at 20:12
@AnuragPallaprolu Double integral = split the integral into two ?? — An SO User, Jul 17 '13 at 20:13
@O.L. Perhaps you could explain briefly why this follows from your link? — user1729, Jul 17 '13 at 20:39
@LittleChild Accepting answers is (of course) entirely up to the OP. However, in the present case, since you admitted in a comment that the posted answer did not help you, I wonder why you accepted this answer (... 7 minutes after it got posted!). A consequence is that it makes other, perhaps more satisfying, answers less likely to be posted. — Did, Jul 29 '13 at 15:58

Elias Costa · Accepted Answer · 2013-07-17T20:37:05.357

3

Hint: From the viewpoint of improper Lebesgue integrals or in sense of Cauchy principal value is integral is legitimate. Integration by parts. \begin{align} \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x} \mbox{d} x = & \lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\dfrac{\sin(x)}{x} \mbox{d} x + \lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\dfrac{\sin(x)}{x} \mbox{d} x \\ = & \lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\sin(x)(\log x)^\prime \mbox{d} x + \lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\sin(x)(\log x)^\prime\mbox{d} x \\ \end{align}

edited Jul 17 '13 at 20:37

answered Jul 17 '13 at 20:11

Elias Costa

14,658

I don't think you're allowed to take the limit that way. You can't take to 0 and to infinity in the same limit. Perhaps there's a rule that says when that's okay? – Adar Hefer Jul 17 '13 at 20:13
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Yeah, it is bad form to compute these as a single limit. Once you know the integral converges, you can get the answer this way; but, without proving convergence, this can give you unfortunate results in some cases. – Nick Peterson Jul 17 '13 at 20:16
Oh, if I already know convergence then this will reproduce the result had I taken the limit using two limits? – Adar Hefer Jul 17 '13 at 20:20
Well this is a hint. Assuming that the integral converges and proceeding step by step by the definition of improper integral you get these equalities between steps. – Elias Costa Jul 17 '13 at 20:21
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How does this help to go toward the answer? – Sungjin Kim Jul 17 '13 at 20:38
@i707107 uhm.. very little :) – An SO User Jul 18 '13 at 06:26
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@LittleChild: Well.. if this answer helped you very little, then why would you accept it? Using integration by parts on this integral only makes it more complicated. So, I even doubt that this answer helped you "very little". I recommend following the link and read answers there. – Sungjin Kim Jul 18 '13 at 07:22
Since $\frac{\sin x}{x}$ is even, then: $\int_{-\infty}^\infty\frac{\sin x}{x}dx=2\int_0^\infty\frac{\sin x}{x}dx=2\cdot\frac{\pi}{2}=\pi$ – Vessel Mar 11 '20 at 23:49

A 'complicated' integral: $ \int \limits_{-\infty}^{\infty}\frac{\sin(x)}{x}$

1 Answers1